Combustion of glucose $(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6)$ produces $\mathrm{CO}_2$ and water. The amount of oxygen (in $\mathrm{g}$) required for the complete combustion of $900 \mathrm{~g}$ of glucose is :

[Molar mass of glucose in $\mathrm{g} \mathrm{~mol}^{-1}=180$]

<p>To determine the amount of oxygen required for the complete combustion of 900 g of glucose, we need to follow these steps:</p> <p>First, let's write the balanced chemical equation for the combustion of glucose $(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6)$:</p> <p>$$\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 + 6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2 + 6 \mathrm{H}_2 \mathrm{O}$$</p> <p>From the balanced equation, we see that 1 mole of glucose reacts with 6 moles of oxygen $(\mathrm{O}_2)$.</p> <p>Next, let's calculate the number of moles of glucose in 900 g. The molar mass of glucose is given as $180 \ \mathrm{g} \ \mathrm{mol}^{-1}$.</p> <p>Number of moles of glucose = $\frac{900 \ \mathrm{g}}{180 \ \mathrm{g} \ \mathrm{mol}^{-1}} = 5 \ \mathrm{mol}$</p> <p>According to the balanced equation, 1 mole of glucose requires 6 moles of oxygen. Therefore, 5 moles of glucose will require:</p> <p>Number of moles of oxygen = $5 \ \mathrm{mol} \times 6 = 30 \ \mathrm{mol}$</p> <p>Now, we need to find the mass of 30 moles of oxygen. The molar mass of oxygen $\left(\mathrm{O}_2\right)$ is $32 \ \mathrm{g} \ \mathrm{mol}^{-1}$.</p> <p>Mass of oxygen = $30 \ \mathrm{mol} \times 32 \ \mathrm{g} \ \mathrm{mol}^{-1} = 960 \ \mathrm{g}$</p> <p>Therefore, the amount of oxygen required for the complete combustion of 900 g of glucose is <strong>960 g</strong>.</p> <p>So, the correct option is <strong>Option D: 960</strong>.</p>